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Applications of Calculus I

Chemical KineticsThe Derivative as a Function

by Dr. Christian Clausen III

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Derivatives•

Recently in class, you have discussed derivatives.

•

One way to find the derivative of a function is to find the slope of a tangent line.

•

Derivatives (slope of the tangent line) are limits of the average rate of change (slope of the secant line) as the interval gets smaller.

•

Chemists use this approach to analyze data in a subject called chemical kinetics.

2

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Chemical Changes Often Occur at Different Rates•

One major factor is concentration

•

Let’s perform a couple of Hydrogen Explosions to demonstrate this

•

H2 + O2 →H2 O + Boom!!!

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What about reactions in solution? Does concentration have an effect on the rate of change?•

Let’s demonstrate this with a chemical reaction that emits light like a lightning bug.

•

CpdM

+ CpdN

→ CpdZ

+ Light (hν)

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Physical changes of rate

•

Observe my changes in Distance Traveled vs

Time (i.e. rate of change) as I move

across the stage.•

So you can see I will have an average rate of change as I go from one end of the stage to the other

•

But my instantaneous rate of change at any given moment might be different

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x1 x2

( x2, f (x2 ))

( x1, f (x1 )))()( 12 xfxf −

12 xx −

( ) ( )12

12 Change of Rate Averagexx

xfxf−−

==antmsec

Average Rate of Change = Slope of Secant Line

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Instantaneous Rate of Change = Slope of Tangent Line

axafxfmxf

ax −−

==→

)()(lim)('xa

( a, f (a))

x

y Secant Lines

a

( a, f (a))

x

y

Tangent Line

The smaller the interval, the better the average rate of change approximates the instantaneous rate of change

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Rates•

Rates of reactions and unemployment rates both measure a change over time.

•

For example: Problem•

from your text 3.1 (33)

17

t U(t) t U(t)1991 6.8 1996 5.41992 7.5 1997 4.91993 6.9 1998 4.51994 6.1 1999 4.21995 5.6 2000 4.0

This shows the percentage of Americans that were unemployed, U(t), from time=1991 to time= 2000

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Finding the Average Rate of Change of Unemployment

•

The rate at which the unemployment rate is changing, in percent unemployed per year.–

Example:

18

70.01

8.65.7yr.1

)1991()1992()1991(' =−

=−

≈UUU

60.01

5.79.6yr.1

)1992()1993()1992(' −=−

=−

≈UUU

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Approximating the derivative•

A more accurate value approximation to U'(1992) is to take the average between U(1992)-U(1991) and U(1993)-

U(1992).

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05.02

60.070.0)1992(' =−

≈∴U

t U’(t) t U’(t)1991 0.70 1996 –0.351992 0.05 1997 –0.451993 –0.70 1998 –0.351994 –0.65 1999 –0.251995 –0.35 2000 –0.20

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Application to Chemistry

•

Chemists analyze how reaction rates change over time.

•

The derivative of this function (reaction rate) is commonly used in kinetics .

•

To find the derivative, the slope of a tangent line can be used.

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Chemical Kinetics (what is it?)The branch of chemistry that is concerned with the rates (or

speed) of change in the concentration of reactants in a chemical reaction.

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Chemical Kinetics

•

Why study kinetics?–

To determine steps in a chemical reaction

–

To develop a mechanism–

To figure out how and why a reaction occurs

–

Ultimately to learn how to make a reaction go faster or slower

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At UCF, the Chemistry courses that cover kinetics are:

CHM 2046- Fundamentals of Chemistry II; CHM 3411-Physical Chemistry II; CHS 6440- Kinetics and Catalysis

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This is a problem from a CHM 2046 exam that deals with Chemical Kinetics

Decomposition of Hydrogen Peroxide.2H2

O2

(l) → H2

O(l) + O2

(g)

The concentration of H2

O2

changes with time by the following:

Calculate the following:

][)(iondecomposit of Rate 2222 OHk

dtOHd

=−

=

)100.9()100.1()104.5(][ 342822

−−− ×+×−×= ttOH t

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a)

The rate of decomposition of H2

O2 after 10 seconds

b)

The rate constantc)

The [H2

O2

] after 10 seconds

][)(iondecomposit of Rate 2222 OHk

dtOHd

=−

=

)100.9()100.1()104.5(][ 342822

−−− ×+×−×= ttOH t

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Before we can learn about calculus applied to chemical kinetics, we need to know about

Chemical Equations•

In a reaction, reactants are converted to products.

Reactants Products

For example, let’s write a simple acid/base reaction: NaOH (a base) + HCl

(an acid) NaCl (a salt) + H2O (water)

•

The speed of the reaction can be determined by:–

The change of reactants (i.e. NaOH

and/or HCl)

–

The change of products (i.e. NaCl and or H2

O)

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Let’s perform this acid-base reaction with HCl

and NaOH

here in class.

We will see how fast the reaction can occur. The color change will indicate when the reaction is over.

•

“As you can see the reaction is very fast.”•

Too fast to follow the change visually but with the proper instrumentation, we can follow changes in concentration vs

time

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When we gather concentration vs time data, we plot it.

•

Change in time–

Denoted as Δtime (x-axis)

•

Change in concentrationDenoted as: Δ

[reactants ] or Δ [products]

(y-axis)

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Reactants to Products

0102030405060

0 20 40 60 80

Time (s)

Num

ber o

f Mol

ecul

es

ProductReactant

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0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2

Time(s)

conc

entr

atio

n A

BC2

C1

t2 t1

A plot of concentration vs

time shows how the Rate of Reaction is changing

•

The plot is not linear–

The rates change over time

•

The average rate over the time interval t1

to t2

is the change of [reactants] from c1

to c2

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tc = AB line of slope = = rate Average

12

12

ΔΔ

−−

ttcc

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Rate of Reaction

•

NOTE: Slope of [reactants] vs. time is always negative–

Reactants are consumed to form products

–

Notice that the slope of AB is negative

•

Rate must always be expressed as positive numbers

•

To ensure this, use the absolute value

–

Rate calculated from [reactants] =

29

tC

ΔΔ−

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•

Aspirin (acetylsalicylic acid) reacts with water to produce salicylic acid and acetic acid

–

The reaction occurs in your stomach and is called an hydrolysis reaction

–

Salicylic acid is the actual pain reliever and fever reducer–

The reaction was stopped at various points so the concentration of the reactant and product could be observed

Let’s look at an example reaction that all of you experienced-the use of aspirin

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Table 1. Data for the hydrolysis of Aspirin in aqueous solution at pH 7 and 37oC

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Time (h) [aspirin] [Salicylic acid]0 5.55 X 10-3 0 X 10-3

20 5.15 X 10-3 0.40 X 10-3

50 4.61 X 10-3 0.94 X 10-3

100 3.83 X 10-3 1.72 X 10-3

200 2.64 X 10-3 2.91 X 10-3

300 1.82 X 10-3 3.73 X 10-3

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Hydrolysis of AspirinPurple data markers, lines and shadingare for reactantsGreen are for products

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Hydrolysis of Aspirin (rate in terms of the product: salicylic acid)

[ ] [ ]=

−−

=−= hhcidsalicylicacidsalicylicarate hht 00.2

02)00.2(

33

•

We can find the average reaction rate using the reactants or the products.

• For example:

Using salicylic acid from t =0h to t =2.0h

hMh

MM /1022

010040.0 53

−−

×=−×

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Hydrolysis of Aspirin (rate in terms of the reactant: aspirin)•

You may also look at the Aspirin to get reaction rate

•

For example:Use data for Aspirin from t=0h to t=2h

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M/h100.2h0h0.2

M1055.5M1051.5 533

−−−

×=−

×−×-

[ ]=

−−

−=−= h0.0h0.2][][ 02

)h0h2(aspirinaspirinrate t

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Hydrolysis of Aspirin (near the end of the reaction)•

Towards the end of the reaction, we can figure out how the rate has changed

•

For example:The rate found by the salicylic acid

from t=200 to t=300

35

hMh

MM /102.8100

1091.21073.3 633

−−−

×=×−×

[ ]=

−−

−=−= h200h300] [] [ 02

)h300h200(acidsalicyclicacidsalicyclicrate t

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ANSWER •

Using the data for Aspirin in Table 1, determine the rate of the reaction from t=200 h to t=300 h

37

[ ] [ ]=

−−

==−= hhaspirinaspirin

rate tt 200300200300

)300200(

hMh

MM /102.8100

1064.21082.1 633

−−−

×=×−×

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Rates: average vs. instantaneous•

Average rates are over a period of time–

Gives limited information

•

As the time intervals get smaller and smaller, they approach a particular “instance”

•

The concentration at that point vs. time is called the instantaneous rate

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YOU WILL CALCULATE THE INSTANTANEOUS RATE OF CHANGE OF A FUNCTION BY USING A NON-GRAPHICAL PROCEDURE

* Remember that the Instantaneous Rate is when change →0 and Δtime →0

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Finding instantaneous rateInstantaneous rate:When the Δc 0 and Δt 0Slope of tangent, EF, hits the

curve at c’

and t’

(-slope tangent) = instantaneous rate

Since the tangent has a negative slope, you must use a negative sign to express the rate as a positive number.

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The initial rate of the reaction

•

Is very important, especially in complex reactions

•

Is at the start of the reaction

•

Is the line of steepest slope–

Fastest rate

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TimeC

once

ntra

tion